Question: What is the slope of the line tangent to $f(x) = 2x^{2}-4x-3$ at $x = 1$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x+\Delta x)^{2}-4(x+\Delta x)-3) - (2x^{2}-4x-3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x^{2}+2x \Delta x+\Delta x^{2})-4(x+\Delta x)-3) - (2x^{2}-4x-3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2x^{2}+4(x \Delta x)+2\Delta x^{2}-4x-4(\Delta x)-3-2x^{2}+4x+3}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{4(x \Delta x)+2\Delta x^{2}-4(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} 4x+2(\Delta x)-4$ $ = 4x-4$ $ = (4)(1)-4$ $ = 0$